Math puzzles!

[shouri]

Hi guys. I'm here to challenge all the mathy/puzzle loving people of SP. Even if you don't consider yourself a part of this category, give these a shot! You never know, you might see something they don't :D



Let's begin!

1) What two numbers, that between them contain the digits 1 through 9, have the largest product when multiplied?

ex: 12345*6789 are two numbers that multiply to 83810205, but you can do better than that! As you can see I used each of the nine digits only once!



2) Use the numbers 987654321 in that order. Add in any + - signs where ever is needed to make an expression that totals to 100.

ex 9+8+7+654-321=(not 100). Your job is to find an expression that does. But you MUST keep the numbers in that order. Multiple solutions.



(I'll add in more later. Try not to cheat please :O It ruins the fun of the people trying these.)

[Jon]

1) 9765*84321 = 823394565
2) (9*8)+7+6+5+4+3+2+1

[shouri]

1) 9765*84321 = 823394565
Nope. It's possible to go about 20 million higher than that. Give others a chance before you answer again :D
2) (9*8)+7+6+5+4+3+2+1
Good Good. People can keep submitting more.

[Locked]

9654 * 87321 = 842996934

[Shidoshi]

1) 94321*8765 = 826723565

Little bit higher, by 3 million.

ninja'd with a better answer.

[Holypie]

97531*8462=842,862,902

[Shidoshi]

Little bit of trying around Excel:

2) 98-76+54+3+21 (thought it was just + or - signs, what Jon did was cheating sorta).

[shouri]

9654 * 87321 = 842996934
Closest answer so far.

Give up yet? :D

[Shidoshi]

My second try here:

1) 9643*87521 = 843 965 003

[Locked]

Closest answer so far.

Give up yet? :D

8754 * 96321 = 843194034

[Holypie]

1) 9642*87531=843 973 902

Slightly better

[shouri]

1) 9642*87531=843 973 902

Slightly better

Ding ding ding. :D

Okay next up:

3) Find three three-digit numbers of the form n, 2n, and 3n such that, collectively they contain the numbers from 1 through 9 exactly once each.

ex: Here's one: 192 384 576. There's three more such trios.

4)
(ignore the ~, I just need it so the problem lines up)
~SEND
+MORE
MONEY

Each different letter represents a unique digit. Which ones are they?

[Jon]

Little bit of trying around Excel:

2) 98-76+54+3+21 (thought it was just + or - signs, what Jon did was cheating sorta).
Yeah, I thought about it while afk after I posted it and remembered it was supposed to be +/-. :/ I'll do a legit one, here:

2) 98-7+6-5+4+3+2-1


3) 219 438 657
4) I don't understand what I'm trying to achieve here. Ah. After Googling, an image of the equation (http://i.imgur.com/NK5XP.png) cleared it up for me. But the page also gave the answer away.

[Locked]

273 546 819
327 654 981
219 438 657
192 384 576

I don't think number 4 is fair because I've done it before and is fairly common.

~9567
+1085
------
10652

[shouri]

273 546 819
327 654 981
219 438 657
192 384 576

I don't think number 4 is fair because I've done it before and is fairly common.

~9567
+1085
------
10652

Well the math puzzles I'll give out will be a mix of common with... wait wut? style problems (as seen in the n, 2n, 3n problem).

5) This is a gambling game problem:

Players bet on a number between 1 and 6 (including 1 and 6). The operator (think dealer) tosses three dice. If all three dice show the same face value, say three 6s, those who bet on 6 win three times their bet, and everyone else loses. If the dice came up with two 6s and a 3, those who backed 6 win twice their bet, and those who bet on 3 win the amount of their bet (aka, they get their money back AND they get that much more back).

"Since each face has a one in six chance of coming up, when three dice are tossed there are therefore 3 in 6 chances that the number I bet on will appear... and if my number comes up more than once... I win even more! So it must be a good game to play."

ex: you bet $1 on 6.
Triple sixes occur. you get your dollar back and win 3 dollars on top of that.
double sixes occur. you get your dollar back and win 2 dollars on top of that.
single six occurs. you get your dollar back and win 1 dollar on top of that.
no sixes occur. You lose your dollar.


Is it? Your job is to work out the true odds of this game!

[Shidoshi]

Mathematical Hope for the value returned for this game is 0.5 of the initial bet:

(5/6)^3 chance to get zero - ~58%
3 * 1/6 * (5/6)^2 chance to get what you bet - ~35%
3 * (1/6)^2 * 5/6 chance to get double your bet ~7%
(1/6)^3 chance to get triple your bet ~0.5%

It adds to exactly 100% and multiplying it by the returns will give you 0.5 =D

It means that on average you will walk off with half of what you invested.
The statement that "Since each face has a one in six chance of coming up, when three dice are tossed there are therefore 3 in 6 chances that the number I bet on will appear..." is already wrong. The chance that at least one face will be the one you chose is (1 - (5/6)^3) = ~42%.

[shouri]

Whoops, I realized that I misworded it, so I added in an example bet with payouts. Sorry shidoshi D:

[chrome]

P(r) = C(n,r)*(p^r)*[q^(n-r)]
p = 0.167
q = 0.833
n = 3
r = # times a given face of a die comes up

P(r=0) = 0.579
P(r=1) = 0.348
P(r=2) = 0.0697
P(r=3) = 0.00466

assuming win = p(r>1):

p = 0.348 + 0.0697 + 0.00466
= 0.42

avg. gross gain given win (where bet = x) [μ(gain|p(r>1)]: (2x+3x+4x)/3 = 3x

μ = 3x(0.421) - -x(0.579)
= 0.684; you'll walk away with 0.684 times what you bet on average

[XTOTHEL]

6 x 6 x 6 = 216 total combos
5 x 5 x 5 = 125 combos without the number YOU guessed.

125/216 = 57.9% chance that you lose.

[shouri]

6 x 6 x 6 = 216 total combos
5 x 5 x 5 = 125 combos without the number YOU guessed.

125/216 = 57.9% chance that you lose.

That's only the first part. You have the losses part right, now you have to split up the winning sections based off expected gain. Do the gain amounts outweigh the loss chance?

Also, not yet Chrome D:

[XTOTHEL]

6 x 6 x 6 = 216 total combos
5 x 5 x 5 = 125 combos without the number YOU guessed.

125/216 = 57.9% chance that you lose.

ok...

67/216 combos that you get 1x, 31.0%
18/216 combos that you get 2x, - 8.33%
6/216 combos that you get 3x - 2.78%

1 x 0.31 + 2x 0.0733 + 3x0.0278 = 0.54

[shouri]

ok...
125/216 = 57.9%
67/216 combos that you get 1x, 31.0%
18/216 combos that you get 2x, - 8.33%
6/216 combos that you get 3x - 2.78%


1 x 0.31 + 2x 0.0733 + 3x0.0278 = 0.54


bolded numbers are incorrect.... also, you forgot to include something in that last equation.... loses have to be taken into account too... not just wins.

[Kalovale]

In all the likelihood of winning, 1/36 nets you 3x bet, 2*(5/36) nets you 2x bet and the rest nets you 1x bet.
On average: 3/36 + 20/36 + 25/36 = 1.33x the bet money.

With a 42% chance to win at 133% bet money, on average you win 42%*133% = 55.9% of the bet money, and lose 57.9% of the bet money per bet.

Not entirely sure I'm not speaking nonsense here, but whatever.

[shouri]

In all the likelihood of winning, 1/36 nets you 3x bet, 2*(5/36) nets you 2x bet and the rest nets you 1x bet.
On average: 3/36 + 20/36 + 25/36 = 1.33x the bet money.

With a 42% chance to win at 133% bet money, on average you win 42%*133% = 55.9% of the bet money, and lose 57.9% of the bet money per bet.

Not entirely sure I'm not speaking nonsense here, but whatever.

Lol, take your time next time.... you can't win 55.9% and lose 57.9% of your money xD

it's 1/216 gets you 3x bet (other errors too).

[Kalovale]

Oh, there's the possibility of winning the other dice when you get double-win too. Hmm..

And I don't see why I can't win those amounts on average.

[shouri]

Also, the rest doesn't net you 1x bet.... you sure CAN get 1 x bet... but you can also lose your bet... meaning -1x bet

[Kalovale]

Also, the rest doesn't net you 1x bet.... you sure CAN get 1 x bet... but you can also lose your bet... meaning -1x bet

I was accessing only winning bets.

[shouri]

I was accessing only winning bets.

yeah, noticed that after. too lazy to edit xD

[Kalovale]

Just to clarify, one correct guess will net you 1x the bet amount, regardless of combo, correct?
E.g: A 6-6-6 guess with 6-6-6 results will gain as much as a 6-6-3 guess with 6-6-3 results.

Silly me, it doesn't matter.

[Mibs]

You people are crazy lol.
I wish I was good enough at math for this.

[shouri]

You people are crazy lol.
I wish I was good enough at math for this.

I didn't learn to do this 'til I got to college (never took stats)... but plenty of people learn to do something real close to this during high school D:

[Shidoshi]

(5/6)^3 chance to get (-1) - ~58%
3 * 1/6 * (5/6)^2 chance to get (+1) - ~35%
3 * (1/6)^2 * 5/6 chance to get (+2) ~7%
(1/6)^3 chance to get (+3) ~0.5%

Averages to -0,0787

[shouri]

(5/6)^3 chance to get (-1) - ~58%
3 * 1/6 * (5/6)^2 chance to get (+1) - ~35%
3 * (1/6)^2 * 5/6 chance to get (+2) ~7%
(1/6)^3 chance to get (+3) ~0.5%

Averages to -0,0787

Love the avatar btw... and yes you're correct.

When betting a dollar you're only receiving back about 92 cents (aka losing out on about 8 cents per game). Let's say six people play a minute a game, with a house minimum of $5 per play... the even though they only lose out on 8% per play... the house gets to make over $100 an hour off of them :D



6) Another gambling game! (probably a bit easier this time)
Let's check the license plates of the next twenty cars we see and notice the last two digits of each plate. I'll bet you $20 that at least two of the cars will have plates with the same last two digits. So you win if there is no match... you lose if there is a match.

"Hmm.... since we're worried about the last two digits... there are 100 choices ranging from 00 to 99. The chance of two of them matching is 1/100 for two chars... so since there's 20 cars, there's about a 20/100 chance that they make.... or 80% chance I'll win...sweet... let's start."

ex Plate1: F567RET plate2: E967WER match up. (Just pretend ALL plates are like this "letter-number-number-number-letter-letter-letter", ignoring custom plates.)

The bet is $20 at 1:1 (aka if you win, you get $20, if you lose you lose $20). This may or may not be bad odds to play at.... what if you were offered 5:1 (aka a bet of $20 can net you $100 if you win).

[nRxUs]

What's the question? Whether you should play?
If that's the question then of course not, you have to take into account permutations and make a binomial distribution which shows that the chances of you losing are over .9.
Similar to that question of if you have at least 30 people in the room there is a huge chance that at least two of them share the same birthday. Permutations win.

[Shidoshi]

The chance that all twenty car plates will be different is:
100!/((100 - 20)! * (100^20)) = 13,04%

Or the complementary, the chance that at least one plate will be the same is 86,96%

So the odds are extremely against the guy betting they'll all be different (even a 1:5 bet is disadvantageous).
Not gonna calculate the expected value here, it's not hard though.

For a group of N possibilities, the chance that in n different picks all will be different is:

N!/((N - n)! * N^n))

[shouri]

What's the question? Whether you should play?
If that's the question then of course not, you have to take into account permutations and make a binomial distribution which shows that the chances of you losing are over .9.
Similar to that question of if you have at least 30 people in the room there is a huge chance that at least two of them share the same birthday. Permutations win.

Chance of losing is NOT over .9 actually. Also remember, the pay out is IMPORTANT. If you had a 90/10 chance (only 10% chance of winning)... but the pay out was higher than 10:1 it'd be worth playing... even with the high chance of losing.

The questions are as follows:

is it worth it to play at a 1:1 payout? You pretty much answered this part. (aka, no)
is it worth it to play at a 5:1 payout? This one's a bit trickier. You'll have to calculate the EXACT percentages to see if 5:1 is a good payout or not.

[Kalovale]

Lol, that dice problem, I went from guessing one dice to guessing three dice without even knowing what happened.

[shouri]

The chance that all twenty car plates will be different is:
100!/((100 - 20)! * (100^20)) = 13,04%

Or the complementary, the chance that at least one plate will be the same is 86,96%

So the odds are extremely against the guy betting they'll all be different (even a 1:5 bet is disadvantageous).
Not gonna calculate the expected value here, it's not hard though.

For a group of N possibilities, the chance that in n different picks all will be different is:

N!/((N - n)! * N^n))

Good good. Since you have the percentages already... the expected value is -1*86.96 + 1*13.04=-73.92 cents, if the pay out is 1:1. even at 5:1 the pay out is -1*86.96 + 5*13.04=-21.76... you'd need 7:1 payout for the game to be worth it for you.


Time for some darts trivia/problems


For reference, in darts, you throw darts at a circular board sectioned off into 20 pieces... think of cutting cake into 20 slices. The numbers 1 through 20 are mixed through the "slices" and landing there gives you that many points. There are also small sections which double or triple the score. So it's possible to get triple 11, for 33 points in one shot. The very center is a circle with another circle around it. The very center is worth 50 points, and the small ring around it is 25 points.

http://upload.wikimedia.org/wikipedia/commons/thumb/4/42/Dartboard.svg/250px-Dartboard.svg.png


The goal in the particular game I'll ask about is to start off at a score (say 300) and work your way down to exactly zero. If you hit a 20, your score drops from 300 to 280. If you're at 5 and hit a 20, you stay at 5 (no going into negatives).

7) What's the lowest score that cannot be scored with one dart throw?
8) What is the lowest score that you can be left with where the game CANNOT be finished in two darts?
9) What is the lowest score that you can be left with where the game CANNOT be finished in THREE darts?
10) Is it possible to score 100 points using three triples? what about four triples?
11) There are four ways to score 26 points using a triple then a double... name them. Ex: triple 1 and double 6 gets you 15. But you must find 4 ways to get a 26 this way.



Try and answer only 1 question at a time to let others have a shot at solving 'em. Also, number your response please :D

[rethic]

7) 23?

[ShinkuDragon]

11:
hit 10*2 and 2*3
7*2 and 4*3
4*2 and 6*3
1*2 and 8*3

woo, new title
<-

[shouri]

7) 23?


11:
hit 10*2 and 2*3
7*2 and 4*3
4*2 and 6*3
1*2 and 8*3

woo, new title
<-

Yes and yes :D

If no one answers 8 through 10 by 8 pm Pacific... you guys can go ahead and answer 1 more each :3

[shouri]

Okay guys, no more answers so far~

Go ahead and have at em ( 7 through 10 )

[ShinkuDragon]

Okay guys, no more answers so far~

Go ahead and have at em ( 8 through 10 )

fixed, and i'm stumped here, lol, don't feel like doing them one at a time atm lol

[Shidoshi]

8) 103 but that took me a LOOOOOOOOOOOOONG time to find out. I just went through most 1-1 combos possible from 80 to 103 (since numbers before 80 are easy to make).

10) It's impossible to make a number that's not multiple of 3 with 3 numbers that are multiples of 3.
3n + 3m + 3k = 100, where n, m and k are integers.
n + m + k = 100/3 the sum of three integers can't be equal to a rational number.

For 4 triples the same reasoning goes,
3n + 3m + 3k + 3p = 100
n + m + k + p = 100/3, the sum of 4 integers can't be equal to a rational number.

[shouri]

8) 103 but that took me a LOOOOOOOOOOOOONG time to find out. I just went through most 1-1 combos possible from 80 to 103 (since numbers before 80 are easy to make).
nope, and i just realized WHY... lololol... in darts it's REQUIRED that your very last shot is a double, no singles or triples. So you have to finish single/double, double/double, or triple double
10) It's impossible to make a number that's not multiple of 3 with 3 numbers that are multiples of 3.
3n + 3m + 3k = 100, where n, m and k are integers.
n + m + k = 100/3 the sum of three integers can't be equal to a rational number.
good good :D
For 4 triples the same reasoning goes,
3n + 3m + 3k + 3p = 100
n + m + k + p = 100/3, the sum of 4 integers can't be equal to a rational number.

8&9 still up for grabs :3

rofl, my bad guys.. read the bolded

[ShinkuDragon]

8&9 still up for grabs :3

rofl, my bad guys.. read the bolded...judging by those rules and what you posted:

8) 1. you can't win
9) 1. you can't win.

there's got to be something else you forgotten though, otherwise this is a "did you read the notes?" question

[HellenzSin]

9)163
1-20: 2 miss, 1 dart hits 1-20
21-40: 1 miss, 1 dart hits 1-20, 1 dart hits 1-20
41-60: 1 dart hits 1-20, 1 dart hits 1-20, 1 dart hits 1-20
61-80: 1 miss, 1 dart hits 1-20, 1 dart hits 3x 1-20
81-100: 1 dart hits 20, 1 dart hits 1-20, 1 dart hits 3x 20
101-120: 1 dart hits 3x 20, 1 dart hits 2x 20, 1 dart hits 1-20
121-140: 1 dart hits 3x 20, 1 dart hits 3x 20, 1 dart hits 1-20
141: 1 dart hits 3x 20, 1 dart hits 3x 20, 1 dart hits 3x 7
142: 1 dart hits 3x 20, 1 dart hits 3x 20, 1 dart hits 2x 11
143: 1 dart hits 50, 1 dart hits 3x 20, 1 dart hits 3x 11
144: 1 dart hits 3x 20, 1 dart hits 3x 20, 1 dart hits 3x 8
145: 1 dart hits 3x 15, 1 dart hits 50, 1 dart hits 50
146: 1 dart hits 3x20, 1 dart hits 3x20, 1 dart hits 2x 13
147: 1 1 dart hits 3x 20, 1 dart hits 3x 20, 1 dart hits 3x 9
148: 1 1 dart hits 3x 20, 1 dart hits 3x 20, 1 dart hits 2x 14
149: 1 dart hits 3x 13, 1 dart hits 50, 1 dart hits 3x 20
150: 3 darts hit 50
151: 1 dart hits 50, 1 dart hits 3x 17, 1 dart hits 50
152: 1 dart hits 3x 20, 1 dart hits 3x 20, 1 dart hits 2x 16
153: 1 dart hits 3x20, 1 dart hits 3x20, 1 dart hits 3x 11
154: 1 dart hits 3x 20, 1 dart hits 3x 20, 1 dart hits 2x 17
155: 1 dart hits 3x 20, 1 dart hits 50, 1 dart hits 3x 15
156: 1 dart hit 3x 20, 1 dart hits 3x 20, 1 dart hits 3x 12
157: 1 dart hits 50, 1 dart hits 3x 19, 1 dart hits 50
158: 1 dart hits 3x 20, 1 dart hits 50, 1 dart hits 3x 16
159: 1 dart hits 3x20, 1 dart 3x20, 1 dart hits 3x 13
160: 1 dart hits 50, 1 dart hits 50, 1 dart hits 3x 20
161: 1 dart hits 3x20, 1 dart hits 50, 1 dart hits 3x 17
162: 1 dart hits 3x20, 1 dart hits 3x20, 1 dart hits 3x 14
163: can't. You can do everything with combinations of 20 and 50s, but this one. 2 darts 3x 20s, 43 left can't be done, prime. 2x 50s, 63 left, can't be done coz there is no 21. 3x 20 + 50, 53 left can't be done, prime.

[shouri]

...judging by those rules and what you posted:

8) 1. you can't win
9) 1. you can't win.

there's got to be something else you forgotten though, otherwise this is a "did you read the notes?" question

T.T besides 1. Geez I'm really out of it. I keep forgetting to mention stuff D:



9)163
1-20: 2 miss, 1 dart hits 1-20
21-40: 1 miss, 1 dart hits 1-20, 1 dart hits 1-20
41-60: 1 dart hits 1-20, 1 dart hits 1-20, 1 dart hits 1-20
61-80: 1 miss, 1 dart hits 1-20, 1 dart hits 3x 1-20
81-100: 1 dart hits 20, 1 dart hits 1-20, 1 dart hits 3x 20
101-120: 1 dart hits 3x 20, 1 dart hits 2x 20, 1 dart hits 1-20
121-140: 1 dart hits 3x 20, 1 dart hits 3x 20, 1 dart hits 1-20
141: 1 dart hits 3x 20, 1 dart hits 3x 20, 1 dart hits 3x 7
142: 1 dart hits 3x 20, 1 dart hits 3x 20, 1 dart hits 2x 11
143: 1 dart hits 50, 1 dart hits 3x 20, 1 dart hits 3x 11
144: 1 dart hits 3x 20, 1 dart hits 3x 20, 1 dart hits 3x 8
145: 1 dart hits 3x 15, 1 dart hits 50, 1 dart hits 50
146: 1 dart hits 3x20, 1 dart hits 3x20, 1 dart hits 2x 13
147: 1 1 dart hits 3x 20, 1 dart hits 3x 20, 1 dart hits 3x 9
148: 1 1 dart hits 3x 20, 1 dart hits 3x 20, 1 dart hits 2x 14
149: 1 dart hits 3x 13, 1 dart hits 50, 1 dart hits 3x 20
150: 3 darts hit 50
151: 1 dart hits 50, 1 dart hits 3x 17, 1 dart hits 50
152: 1 dart hits 3x 20, 1 dart hits 3x 20, 1 dart hits 2x 16
153: 1 dart hits 3x20, 1 dart hits 3x20, 1 dart hits 3x 11
154: 1 dart hits 3x 20, 1 dart hits 3x 20, 1 dart hits 2x 17
155: 1 dart hits 3x 20, 1 dart hits 50, 1 dart hits 3x 15
156: 1 dart hit 3x 20, 1 dart hits 3x 20, 1 dart hits 3x 12
157: 1 dart hits 50, 1 dart hits 3x 19, 1 dart hits 50
158: 1 dart hits 3x 20, 1 dart hits 50, 1 dart hits 3x 16
159: 1 dart hits 3x20, 1 dart 3x20, 1 dart hits 3x 13
160: 1 dart hits 50, 1 dart hits 50, 1 dart hits 3x 20
161: 1 dart hits 3x20, 1 dart hits 50, 1 dart hits 3x 17
162: 1 dart hits 3x20, 1 dart hits 3x20, 1 dart hits 3x 14
163: can't. You can do everything with combinations of 20 and 50s, but this one. 2 darts 3x 20s, 43 left can't be done, prime. 2x 50s, 63 left, can't be done coz there is no 21. 3x 20 + 50, 53 left can't be done, prime.

You also missed the bolded where you're required to end in a double.



Here: official rules


Double-out: Players must reach "0" exactly by scoring in a double point value segment.

With 32 points to go, a double 16 will end the game. A single 16 will still leave 16 to go, which can be gotten by throwing a double 8, and so on down the line. If an odd number is thrown, another odd number is needed to get back to a double possibility. If one less or one more than the exact score is hit, the player must assume the same score he had at the end of his last turn. (This is called busting.) The rest of the rules hold.
aka, if you get down to 1 point... you bust and have to go back to the previous score you had.

[Riyuran]

8) 99

2-41: 1/miss + 2x 1-20
42-60: 2-20 + 2x 20
61: 3x 7 + 2x 20
62: 2x 11 + 2x 20
63: 3x 13 + 2x 12
64: 2x 20 + 2x 12
65: 3x 13 + 2x 13
66: 3x 14 + 2x 12
67: 3x 19 + 2x 5
68: 3x 14 + 2x 13
69: 3x 15 + 2x 12
70: 3x 20 + 2x 5
71: 3x 15 + 2x 13
72: 3x 16 + 2x 12
73: 3x 19 + 2x 8
74: 3x 16 + 2x 13
75: 3x 15 + 2x 15
76: 3x 20 + 2x 8
77: 3x 17 + 2x 13
78: 3x 16 + 2x 15
79: 3x 19 + 2x 11
80: 3x 20 + 2x 10
81: 3x 17 + 2x 15
82: 3x 20 + 2x 11
83: 3x 19 + 2x 13
84: 3x 18 + 2x 15
85: 3x 15 + 2x 20
86: 3x 20 + 2x 13
87: 3x 19 + 2x 15
88: 3x 16 + 2x 20
89: 3x 19 + 2x 16
90: 3x 20 + 2x 15
91: 3x 17 + 2x 20
92: 3x 20 + 2x 16
93: 3x 19 + 2x 18
94: 3x 18 + 2x 20
95: 3x 19 + 2x 19
96: 3x 20 + 2x 18
97: 3x 19 + 2x 20
98: 3x 20 + 2x 19
99: Not possible! The highest number you can go with the second shot is 2x 20 = 40. So you need to get to 40 (or a lower even number) after the first shot. However, hitting 59 (or a higher odd number) with one shot isn't possible.
100: 3x 20 + 2x 20
101+: Not possible with only 2 darts.

[shouri]

8) 99

2-41: 1/miss + 2x 1-20
42-60: 2-20 + 2x 20
61: 3x 7 + 2x 20
62: 2x 11 + 2x 20
63: 3x 13 + 2x 12
64: 2x 20 + 2x 12
65: 3x 13 + 2x 13
66: 3x 14 + 2x 12
67: 3x 19 + 2x 5
68: 3x 14 + 2x 13
69: 3x 15 + 2x 12
70: 3x 20 + 2x 5
71: 3x 15 + 2x 13
72: 3x 16 + 2x 12
73: 3x 19 + 2x 8
74: 3x 16 + 2x 13
75: 3x 15 + 2x 15
76: 3x 20 + 2x 8
77: 3x 17 + 2x 13
78: 3x 16 + 2x 15
79: 3x 19 + 2x 11
80: 3x 20 + 2x 10
81: 3x 17 + 2x 15
82: 3x 20 + 2x 11
83: 3x 19 + 2x 13
84: 3x 18 + 2x 15
85: 3x 15 + 2x 20
86: 3x 20 + 2x 13
87: 3x 19 + 2x 15
88: 3x 16 + 2x 20
89: 3x 19 + 2x 16
90: 3x 20 + 2x 15
91: 3x 17 + 2x 20
92: 3x 20 + 2x 16
93: 3x 19 + 2x 18
94: 3x 18 + 2x 20
95: 3x 19 + 2x 19
96: 3x 20 + 2x 18
97: 3x 19 + 2x 20
98: 3x 20 + 2x 19
99: Not possible! The highest number you can go with the second shot is 2x 20 = 40. So you need to get to 40 (or a lower even number) after the first shot. However, hitting 59 (or a higher odd number) with one shot isn't possible.
100: 3x 20 + 2x 20
101+: Not possible with only 2 darts.

Excellent, 99 is the correct answer :D

Now do it with three darts :D (At the very least you already know what's possible with 2 darts, so no need to check those :D

[ShinkuDragon]

Excellent, 99 is the correct answer :D

Now do it with three darts :D (At the very least you already know what's possible with 2 darts, so no need to check those :D

imma throw a dart in the dark and say 159

[shouri]

imma throw a dart in the dark and say 159


And BULLSEYE! *ba dum tch*
that is indeed correct~ you can just triple 60 your way to anything before 159 by taking 99 and before for the last two darts. Excellent intuition! THIS IS THE TRUE ESSENCE OF MATH MUHAHAHAHAAHA! *cough* ahem~

12) Continue the sequence (If you've seen it, let everyone else suffer):

[just give me the next line]
1
11
21
1211
111221
312211
...?

[ShinkuDragon]

And BULLSEYE! *ba dum tch*
that is indeed correct~ you can just triple 60 your way to anything before 159 by taking 99 and before for the last two darts. Excellent intuition! THIS IS THE TRUE ESSENCE OF MATH MUHAHAHAHAAHA! *cough* ahem~


bawss. thanks ryu for doing all the hard work for me <3

and i've seen it, but can't remember exactly what the answer was >.<

[Infernal]

And BULLSEYE! *ba dum tch*
that is indeed correct~ you can just triple 60 your way to anything before 159 by taking 99 and before for the last two darts. Excellent intuition! THIS IS THE TRUE ESSENCE OF MATH MUHAHAHAHAAHA! *cough* ahem~

12) Continue the sequence (If you've seen it, let everyone else suffer):

[just give me the next line]
1
11
21
1211
111221
312211
...?

13112221

[shouri]

13112221

Good good, each line just describes the previous one.

In the last line I posted we:

one 3, one 1, two twos, then two ones.
or
13,11,22,21

They're read off by groups not by total.

13) Let's have some fun :3 I would like you guys each to post up a single way to represent pi. Ingenuity wins this one (earlier ones receiving preference... since others will have the chance to see your creativity and just work to out do you... they better outdo you by a LOT to win).... so simple representations or re-writes of well known formulas will be frowned upon. This is a challenge of CREATIVITY. Everyone post your own creative way to represent pi. Just because someone else posted one doesn't mean you can't!

Edit: you do NOT have to represent pi exactly... approximations will do too... so long as they have a few digits of accuracy. As usual, please don't just look them up >.<

[Holypie]

http://www.visitingdc.com/images/arc-de-triomphe-picture.jpg
Sorry, I couldn't help it. I'll do a real one soon.

I'm tossing out all units.
Gravity of Earth's moon (1.67m/s^2) + phi (1.61803399) - Astronomical Unit (1.4 x 10^-1 Tm) = which is 3.14 when rounded to 2 decimal places.

Not very creative, but it's the best I can do on short notice.

[ShinkuDragon]

http://chzmemebase.files.wordpress.com/2010/11/9e7c48aa-1823-4d5f-aa13-40699c72d508.jpg

i... i just had to...

[Shidoshi]

Perimeter of an n sided polygon with a diameter of 1.

http://i.imgur.com/HEIFh.png

The size of the side of this polygon can be calculated as 2*sin(360ş/2n)*(1/2) = sin(180ş/n), that is taking pitagoras on half of a pizza slice of the polygon.
The total perimeter of this polygon is n * sin(180ş/n)

If we take the limit of n tending to infinity we have pi.
yay~

[shouri]

Okay, okay~ this is working so far~
Using equations that aren't as well known (physics, holypie)... and limits that also aren't obvious (Shidoshi, love the signature as well btw :3).... granted this shows up on wikipedia... hopefully you just learned it in class or something :D


Let's see some more creativity, also... L> MOAR math people... get that creativity flowing~ You two (shidoshi/holypie) can post one more if you feel like it :D

[Shidoshi]

That expression to get to pi is pretty common place (sorta), as a circle is just a polygon of infnite sides =D
I had solved that limit myself when I was studying calculus.

Meh, if I could remember there was this Fourrier series that you could use to get the value of pi as an infinite sum of sines and cosines, it's been too long though.

I'll see if I can think of something.

[Stereo]

"Since each face has a one in six chance of coming up, when three dice are tossed there are therefore 3 in 6 chances that the number I bet on will appear... and if my number comes up more than once... I win even more! So it must be a good game to play."

ex: you bet $1 on 6.
Triple sixes occur. you get your dollar back and win 3 dollars on top of that.
double sixes occur. you get your dollar back and win 2 dollars on top of that.
single six occurs. you get your dollar back and win 1 dollar on top of that.
no sixes occur. You lose your dollar.


Is it? Your job is to work out the true odds of this game!
Going back to this one, cause it bothered me that nobody solved it this way...
For each correct die, you win $1.
so 1/6*$1 + 1/6*$1 + 1/6*$1 = $0.50 per game
But: If none of the dice are right, you lose $1
so 5/6*5/6*5/6*$-1 = 125/216*$-1 = $-0.5787 per game
Add them up: $-0.0787 per game, or $0.9213 return on $1 investment.

[shouri]

Going back to this one, cause it bothered me that nobody solved it this way...
For each correct die, you win $1.
so 1/6*$1 + 1/6*$1 + 1/6*$1 = $0.50 per game
But: If none of the dice are right, you lose $1
so 5/6*5/6*5/6*$-1 = 125/216*$-1 = $-0.5787 per game
Add them up: $-0.0787 per game, or $0.9213 return on $1 investment.

Quite the nice solution to that problem :D

[Noah]

http://www.wolframalpha.com/input/?i=average+orbit+velocity+of+earth+*+orbital+perio d+for+earth+%2F%28+2+*+average+distance+from+earth +to+sun%29

Well, that was more exact than I thought it would be.

[shouri]

http://www.wolframalpha.com/input/?i=average+orbit+velocity+of+earth+*+orbital+perio d+for+earth+%2F%28+2+*+average+distance+from+earth +to+sun%29

Well, that was more exact than I thought it would be.

My thoughts exactly.

Now noah, how 'bout you give us a nice approximation :3

[Riyuran]

http://mathurl.com/ccotryn.png
Not creative at all.. but I love the Gaussian Integral. (:

[shouri]

Here, howsabout I give you guys a nice example....

It's one of my favorites:

pi: ln(640320^3 + 744)/sqrt(163)

My image of this borked D:


That is approximately pi.. and is accurate to roughly 30 places I think. :3

[rethic]

http://chzmemebase.files.wordpress.com/2010/11/9e7c48aa-1823-4d5f-aa13-40699c72d508.jpg


This has gotten me very intrigued. How does that work out? I'm a little sad that I can't figure out what's wrong with the logic.

[shouri]

This has gotten me very intrigued. How does that work out? I'm a little sad that I can't figure out what's wrong with the logic.

You can calculate the difference in the perimeters at each step... that sequence doesn't converge to zero. It's just the sequence: 4-pi, 4-pi, 4-pi, 4-pi..... so the error is always 4-pi even after you "repeat to infinity". Likewise the process from the 4th to the 5th panel is technically impossible. It'll never be smooth enough using that method. So you can never say that they are equal, even in the limit. It's not like the n-side polygon inside of a circle trick.... where you let n go to infinity and you get the circle. That one DOES smooth out enough.... but it's really hard to explain without getting into slightly complicated math.

[ShinkuDragon]

You can calculate the difference in the perimeters at each step... that sequence doesn't converge to zero. It's just the sequence: 4-pi, 4-pi, 4-pi, 4-pi..... so the error is always 4-pi even after you "repeat to infinity". Likewise the process from the 4th to the 5th panel is technically impossible. It'll never be smooth enough using that method. So you can never say that they are equal, even in the limit. It's not like the n-side polygon inside of a circle trick.... where you let n go to infinity and you get the circle. That one DOES smooth out enough.... but it's really hard to explain without getting into slightly complicated math.which is why this troll was so beautiful, to the point it had to be "solved" by majors at math, so that everyone would stop flipping their shits.

[shouri]

which is why this troll was so beautiful, to the point it had to be "solved" by majors at math, so that everyone would stop flipping their pomegranates.

The "smoothness" is the only tough part to explain. The sequence is pretty easy to explain and suffices to disprove it...

but it's still such a wonderful troll x3



L> MOAR approximations to pi

14) And in the mean time, let's have a bit more fun with pi. My friend (let's call him Mike) has a very unique social security number... 123-45-6789. Quite fun. I told him his social security number is actually found in the decimal expansion of pi. In fact, I can prove it... can you?

Assumptions/facts we need:
a)pi goes on forever
b)pi is normal... never repeats and is, for all intents and purposes, truly random.

[Stereo]

The "smoothness" is the only tough part to explain. The sequence is pretty easy to explain and suffices to disprove it...I don't think it's sufficient, you can't assume from the start what the perimeter is, because it's what you're trying to solve. If pi = 4, then the perimeter of the circle is 2*pi*r = 4.

This assumption is IMO also a problem for the limits of sin/cos, they're defined based on a value of pi, so of course they'll spit out pi if you take sin^-1(1) in a program that knows sin(pi)=1.

[ShinkuDragon]

please don't tell me the troll math took over the thread.

[Holypie]

If pi is infinite and is truly random and doesn't repeat, then statistically any finite string of numbers will appear since there are an infinite number of possibilities. If it's infinite, everything will happen sooner or later.

[shouri]

If pi is infinite and is truly random and doesn't repeat, then statistically any finite string of numbers will appear since there are an infinite number of possibilities. If it's infinite, everything will happen sooner or later.

Good explanation, kinda.... but not a proof :3

[shouri]

If pi is infinite and is truly random and doesn't repeat, then statistically any finite string of numbers will appear since there are an infinite number of possibilities. If it's infinite, everything will happen sooner or later.

Good explanation, kinda.... but not a proof :3

[nRxUs]

Not 100% sure if this is proof but here it goes,
Given a random and infinite string of numbers N. Let's divide N into an infinite number of 9-digit numbers, let's call them Xi, where i from from 1 to infinite.
We now have an infinite number of 9-digit number strings that are all random. Since each string is random, and we have an infinite number of them we have a probability of 1 of at least one being 123456789. As a matter of fact we have an infinite number of 123456789 strings, but that's another story.

[shouri]

Not 100% sure if this is proof but here it goes,
Given a random and infinite string of numbers N. Let's divide N into an infinite number of 9-digit numbers, let's call them Xi, where i from from 1 to infinite.
We now have an infinite number of 9-digit number strings that are all random. Since each string is random, and we have an infinite number of them we have a probability of 1 of at least one being 123456789. As a matter of fact we have an infinite number of 123456789 strings, but that's another story.

You're getting at it... but you have to explain WHY we have a probability of 1... not just "because there's an infinite amount of them"


Here's how (using your notation):

The probability of any Xi being 123456789 is 1/(1,000,000,000). So the probability that it isn't 123456789 is 999,999,999/1,000,000,000.

The probability that they ALL aren't Xi is:

limit x->inf (999,999,999/1,000,000,000)^x

Which, since the fraction is less than 1, is equal to zero.

Thus the probability that at least 1 of them is 123456789 is 1.


You pretty much had it.